Question

How would you solve this equation? Please show steps

Answer 1

Simple....

you have: $3x= \sqrt{12-12x}$

So...first you know you're trying to find x..

But, you have a square root on one side...to remove this you'd have to square both sides--->>>

$(3x)^{2} = \sqrt{12-12x} ^{2}$

Leaving you with....

$9x^{2} =12-12x$

Move the terms over and set it equal to zero.

-->>

$9x^{2}+12x-12=0$

Factor out a 3...

$3( 3x^{2} +4x-4=0)$

What multiplies to -12 and adds to 4?

6*-2=-12

6+-2=4

Leaving you with...

$3(3x-2)(x+2)=0$

Remember you're solving for 0....

3x-2=0

3x-2=0
    +2  +2

3x=2

$\frac{3x}{3} = \frac{2}{3}$

x=$\frac{2}{3}$

Then-->>

x+2=0

x+2=0
  -2   -2

x=-2

Thus, your answer.